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Suppose that every nonempty definable (without parameters) subset of the group $G$ contains an element of finite order.

Must $G$ be elementary equivalent to a countable torsion group?

My first idea is to take the set of definable elements of finite order but the product of two finite order elements need not be of finite order.

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    Regarding your attempt: Actually, for your group $G$, the product of any two finite order definable elements *does* have finite order! This is because if $a$ is definable, then $\{a\}$ is a definable set, and hence $a$ has finite order. So the set of definable elements is equal to the set of definable elements with finite order, and since the product of two definable elements is definable, this set is a subgroup. The problem is that there's no reason to expect this subgroup to be elementarily equivalent to $G$. – 2017-02-13
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    For example, $(\mathbb{Z}/2\mathbb{Z})^\omega$ is a torsion group, so it satisfies your hypotheses, but the only definable element is $e$, since any two non-identity element of this group are conjugate by an automorphism. – 2017-02-13
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    You could try to get around this problem (that the definable elements don't necessarily form an elementary substructure) by Skolemizing your theory. But adding Skolem functions adds new definable sets, and you might lose your hypothesis about definable sets containing elements of finite order. Thus, you would have to be careful to make sure that the Skolem functions always output elements of finite order. These sorts of careful choices are exactly what's accomplished by the Henkin construction proof of omitting types, and you're essentially led to the strategy in my answer after all. – 2017-02-13

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This is a nice application of the omitting types theorem.

Let $\varphi_n(x)$ be the formula $x^n = e$ (where of course $x^n$ is an abbreviation for the $n$-fold product of $x$). Let $\Sigma(x) = \{\lnot \varphi_n(x)\mid n\in \omega\}$. An element satisfies the partial type $\Sigma(x)$ if and only if it has infinite order. So we seek a countable model of $\text{Th}(G)$ in which no element satisfies $\Sigma(x)$.

By the omitting types theorem, such a model exists as long as there is no formula $\psi(x)$ such that $\text{Th}(G)\models \exists x\, \psi(x)$ and $\text{Th}(G)\models \forall x\, \psi(x)\rightarrow \lnot\varphi_n(x)$ for all $n\in \omega$. This is exactly the hypothesis you're given: no nonempty definable subset of $G$ contains only elements of infinite order.

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    Nice! +1. ${}{}{}{}$ – 2017-02-13